How Does Viscous Resistance Affect Object Motion?

[hastings]

Homework Statement

An object with mass m=0.1 kg is thrown with an initial speed v0=20m/s in a viscous matter which opposes its motus with a resistance force of F=-Bv (v is the speed and B=2kg/s). Find the distance this object has covered in the viscous matter. Leave aside gravity force.

Homework Equations

F=m*a

The Attempt at a Solution

I tried this way

\vec F=-\beta \vec v=m \cdot a

\frac{-\beta \vec v}{m}= a= \frac{dv}{dt}=\frac{ds}{d^2t}

\frac{-\beta \vec v}{m} d^2t=ds

\int{\frac{-\beta \vec v}{m} }d^2t=\int ds

\int{\frac{-\beta \vec v}{m} } \cdot t dt=s

\displaystyle{-\frac{1}{2}} \frac{\beta v }{m}t^2+K=s

How to continue? Is it right till this point?

 

[Doc Al]

Realize that v is not a constant but a function of t. Stick with this:
\frac{-\beta \vec v}{m}= a= \frac{dv}{dt}

Rearrange so that v is on one side and t on the other. Then you can integrate.

 

[hastings]

you mean like this?
-\frac{\beta dt}{m}=\frac{1}{v}dv

\int{ -\frac{\beta}{m}}dt=\int \frac{1}{v}dv

-\frac{\beta}{m}t=ln(v)+K

How to get K? Should i substitute t0=0s & v0=20m/s?

 

[hastings]

to get the distance "s", I need a formula with a, t and v
Where do I go with an expression with 2 unknown quantities, v and t?
-\frac{\beta}{m}t=ln(v)-ln20

 

[hastings]

suppose I do ...
-\frac{\beta}{m}t+\log20=\log v

\frac{ds}{dt}=v=e^{-\frac{\beta}{m}t+\log20}

\int{ds}=\int{e^{-\frac{\beta}{m}t+\log20} }dt

\int ds=\int (C \cdot e^{\alpha t})dt \ \ \mbox{where \alpha=\frac{-\beta}{m} } \ \mbox{ and C=e^{\log20} }

s=\frac{C}{\alpha}e^{\alpha t}+K_1

I assume k1=0; I know all the quantities except t

 

[Doc Al]

Looks good to me (with a bit of correction).

You can simplify a bit. For example:
C=e^{\ln20}
What does that equal?

s=\frac{C}{\alpha}e^{\alpha t}+K_1

I assume k1=0; I know all the quantities except t

Realize that k1 cannot equal zero if s = 0 at t = 0.

 

[hastings]

ok, Doc Al. got your message about posting the same topic more than once.
Won't do it again. This is what happened: after waiting a long time I didn't get much help in this section so I tried my luck on some other section - where eventually I got some interesting hint to solve this problem (THANKS A LOT TO "arunbg"!).
Here's what I did

F_R= -\beta \cdot v=m \cdot a

(1) \ a=-\frac{\beta v}{m}

(2) \ a=v \frac{dv}{dx} \ \mbox{ infact } \displaystyle{(v \cdot \frac{ \frac{dv}{dt}}{\frac{dx}{dt} } = v \cdot \frac{a}{v} = a)}

v \frac{dv}{dx}= -\frac{\beta v}{m}

\frac{dv}{dx}= -\frac{\beta}{m}

dv=-\frac{\beta}{m} dx

\int{dv}= \ -\int{\frac{\beta}{m}dx}

v=-\frac{\beta}{m}x

x=-\frac{mv}{\beta}

x=-\frac{0.1 \mbox{ kg } (20 \mbox{ m/s})}{2 \mbox{ kg/s}}=-1m

I'm not sure whether this is the correct way, but the result is 1m.

 

[hastings]

Looks good to me (with a bit of correction).

You can simplify a bit. For example:
C=e^{\ln20}
What does that equal?


Realize that k1 cannot equal zero if s = 0 at t = 0.

C=e^{\ln 20} \ \Longrightarrow C=20?

You are right about k1. with t=0 and x=0 i get

0=\frac{20}{\alpha} e^{\alpha \cdot 0}+K_1

0=\frac{20}{\alpha}+K_1 \ \Longrightarrow K_1=-\frac{20}{\alpha}

 

[hastings]

Suppose I try with the definite integral...

 

[hastings]

Suppose I try with the definite integral...

a=\frac{dv}{dt}=-\frac{\beta v}{m}

\frac{1}{v}dv=-\frac{\beta}{m}dt

\int_{v_0}^v{\frac{1}{v}}dv=-\frac{\beta}{m} \int_0^t{}dt

\ln v -\ln v_0=-\frac{\beta}{m}t

\ln \frac{v}{v_0}=-\frac{\beta}{m}t

\frac{v}{v_0}=e^{-\frac{\beta}{m}t}

v=\frac{dx}{dt}=v_0e^{-\frac{\beta}{m}t}

dx=v_0e^{-\frac{\beta}{m}t}dt

\int_0^x{dx}= \int_0^t {v_0e^{-\frac{\beta}{m}t}}dt

x=v_0[-\frac{m}{\beta}e^{-\frac{\beta}{m}t}]_0^t=v_0[\frac{m}{\beta}-\frac{m}{\beta}e^{-\frac{\beta}{m}t}]= \frac{m v_0}{\beta}[1-e^{-\frac{\beta}{m}t}]

So in the end I get

x=\frac{mv_0}{\beta} \displaystyle{[1-e^{-\beta t/m }]}

How to go foward now? The answer is 1m

 

[hastings]

What I don't know is the instant (time) this object stops or anyway comes out of the viscous medium. Just a step away from the solution...How do I go on now?

Anybody there to help me? Any suggestion is appreciated.

 

[cristo]

ok, Doc Al. got your message about posting the same topic more than once.
Won't do it again. This is what happened: after waiting a long time I didn't get much help in this section so I tried my luck on some other section - where eventually I got some interesting hint to solve this problem (THANKS A LOT TO "arunbg"!).
Here's what I did

F_R= -\beta \cdot v=m \cdot a

(1) \ a=-\frac{\beta v}{m}

(2) \ a=v \frac{dv}{dx} \ \mbox{ infact } \displaystyle{(v \cdot \frac{ \frac{dv}{dt}}{\frac{dx}{dt} } = v \cdot \frac{a}{v} = a)}

v \frac{dv}{dx}= -\frac{\beta v}{m}

\frac{dv}{dx}= -\frac{\beta}{m}

dv=-\frac{\beta}{m} dx

\int{dv}= \ -\int{\frac{\beta}{m}dx}

This method is fine up until you integrate and decide to omit the constant of integration! The next line should read

v=-\frac{\beta}{m}x +A for some constant, A. Use your initial conditions to find A, then proceed as you did to find x.

 

[hastings]

Assuming that "t" is endless (of course not negative, never heard of "t=-4s")

\lim_{t \to +\infty}\frac{mv_0}{\beta} \displaystyle{[1-e^{-\beta t/m }]}= \frac{mv_0}{\beta} = \frac{0.1kg \ (20 \mbox{ m/s})}{2kg/s}=1m

 

[Doc Al]

You seem to be doing the entire problem over and over several ways. Nothing wrong with that as long as you realize that any correct way will give you the answer if you stick with it. You had all you needed for the answer in post #5, once you found the correct constant.

So in the end I get

x=\frac{mv_0}{\beta} \displaystyle{[1-e^{-\beta t/m }]}

How to go foward now? The answer is 1m

Examining this solution should tell you something about how the object will move. The distance reaches a limit as time goes to infinity.

What I don't know is the instant (time) this object stops or anyway comes out of the viscous medium. Just a step away from the solution...How do I go on now?

Take the limit.

Assuming that "t" is endless (of course not negative, never heard of "t=-4s")

\lim_{t \to +\infty}\frac{mv_0}{\beta} \displaystyle{[1-e^{-\beta t/m }]}= \frac{mv_0}{\beta} = \frac{0.1kg \ (20 \mbox{ m/s})}{2kg/s}=1m

There you go.

Just for fun, if you can stand it, plug in a few values of time (1 sec, 2 sec, etc...) to see that you don't really have to wait forever for the object to go pretty much as far as it will go.